Let $f(x)=5x^3-6\cos(x)$. $f'(x)=$
Solution: The expression for $f(x)$ includes $\cos(x)$. Remember that the derivative of $\cos(x)$ is $-\sin(x)$. Put another way, $\dfrac{d}{dx}[\cos(x)]=-\sin(x)$. $\begin{aligned} f'(x)&=\dfrac{d}{dx}[5x^3-6\cos(x)] \\\\ &=5\dfrac{d}{dx}(x^3)-6\dfrac{d}{dx}[\cos(x)] \\\\ &=5\cdot(3x^2)-6(-\sin(x)) \\\\ &=15x^2+6\sin(x) \end{aligned}$ In conclusion, $f'(x)=15x^2+6\sin(x)$